Difference between revisions of "2017 AIME I Problems/Problem 4"
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− | ==Problem | + | ==Problem == |
A pyramid has a triangular base with side lengths <math>20</math>, <math>20</math>, and <math>24</math>. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length <math>25</math>. The volume of the pyramid is <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>. | A pyramid has a triangular base with side lengths <math>20</math>, <math>20</math>, and <math>24</math>. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length <math>25</math>. The volume of the pyramid is <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>. | ||
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<cmath>\overline {OM} + \overline {OC} = \overline {CM} = 16</cmath> | <cmath>\overline {OM} + \overline {OC} = \overline {CM} = 16</cmath> | ||
− | Let <math>\overline {OM} = d</math>. | + | Let <math>\overline {OM} = d</math>. Then <math>OC=OA=\sqrt{d^2+12^2}.</math> |
Equation <math>(1)</math>: | Equation <math>(1)</math>: | ||
<cmath>d + \sqrt {d^2 + 144} = 16</cmath> | <cmath>d + \sqrt {d^2 + 144} = 16</cmath> | ||
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Take a small triangular element in the pyramid. We know that it’s area is proportional to the height from the vertex to the base. Hence, we know that <math>\frac{A_{small element}}{A} = \frac{h^2}{H^2} \implies A_{small element} = \frac{Ah^2}{H^2}</math>. Now integrate it taking the limits <math>0</math> to <math>H</math> | Take a small triangular element in the pyramid. We know that it’s area is proportional to the height from the vertex to the base. Hence, we know that <math>\frac{A_{small element}}{A} = \frac{h^2}{H^2} \implies A_{small element} = \frac{Ah^2}{H^2}</math>. Now integrate it taking the limits <math>0</math> to <math>H</math> | ||
− | ==Shortcut== | + | ===Shortcut=== |
Here is a shortcut for finding the radius <math>R</math> of the circumcenter of <math>\triangle ABC</math>. | Here is a shortcut for finding the radius <math>R</math> of the circumcenter of <math>\triangle ABC</math>. | ||
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Then continue as before to use the Pythagorean Theorem on <math>\triangle AOP</math>, find <math>h</math>, and find the volume of the pyramid. | Then continue as before to use the Pythagorean Theorem on <math>\triangle AOP</math>, find <math>h</math>, and find the volume of the pyramid. | ||
+ | |||
+ | ===Another Shortcut (Extended Law of Sines)=== | ||
+ | Take the base <math>\triangle ABC</math>, where <math>AB = BC = 20</math> and <math>AC = 24</math>. Draw an altitude from <math>B</math> to <math>AC</math> that bisects <math>AC</math> at point <math>D</math>. Then the altitude has length <math>\sqrt{20^2 - 12^2} = \sqrt{16^2} = 16</math>. Next, let <math>\angle BCA = \theta</math>. Then from the right triangle <math>\triangle BDC</math>, <math>\sin \theta = 4/5</math>. From the extended law of sines, the circumradius is <math>20 \cdot \dfrac{5}{4} \cdot \dfrac{1}{2} = \dfrac{25}{2}</math>. | ||
==Solution 2 (Coordinates)== | ==Solution 2 (Coordinates)== | ||
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==Solution 4 (Symmetry)== | ==Solution 4 (Symmetry)== | ||
− | ''' | + | [[File:2017 AIME I 4.png|450px|right]] |
+ | |||
+ | Notation is shown on diagram. | ||
+ | <cmath>AM = MB = c = 12, AC = BC = b = 20,</cmath> | ||
+ | <cmath>DA = DB = DC = a = 25.</cmath> | ||
+ | <cmath>CM = x + y = \sqrt{b^2-c^2} = 16,</cmath> | ||
+ | <cmath>x^2 - y^2 = CD^2 – DM^2 = CD^2 – (BD^2 – BM^2) = c^2 = 144,</cmath> | ||
+ | <cmath>x – y = \frac{x^2 – y^2}{x+y} = \frac {c^2} {16} = 9,</cmath> | ||
+ | <cmath>x = \frac {16 + 9}{2} = \frac {a}{2},</cmath> | ||
+ | <cmath>h = \sqrt{a^2 -\frac{ a^2}{4}} = a \frac {\sqrt{3}}{2},</cmath> | ||
+ | <cmath>V = \frac{h\cdot CM \cdot c}{3}= \frac{16\cdot 25 \sqrt{3} \cdot 12}{3} = 800 \sqrt{3} \implies \boxed {803}.</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 15:55, 5 January 2024
Contents
Problem
A pyramid has a triangular base with side lengths ,
, and
. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length
. The volume of the pyramid is
, where
and
are positive integers, and
is not divisible by the square of any prime. Find
.
Solution
Let the triangular base be , with
. We find that the altitude to side
is
, so the area of
is
.
Let the fourth vertex of the tetrahedron be , and let the midpoint of
be
. Since
is equidistant from
,
, and
, the line through
perpendicular to the plane of
will pass through the circumcenter of
, which we will call
. Note that
is equidistant from each of
,
, and
. Then,
Let . Then
Equation
:
Squaring both sides, we have
Substituting with equation :
We now find that .
Let the distance . Using the Pythagorean Theorem on triangle
,
, or
(all three are congruent by SSS):
Finally, by the formula for volume of a pyramid,
This simplifies to
, so
.
NOTE : If you don’t know or remember the formula for the volume of a triangular pyramid, you can derive it using calculus as follows :
Take a small triangular element in the pyramid. We know that it’s area is proportional to the height from the vertex to the base. Hence, we know that . Now integrate it taking the limits
to
Shortcut
Here is a shortcut for finding the radius of the circumcenter of
.
As before, we find that the foot of the altitude from lands on the circumcenter of
. Let
,
, and
.
Then we write the area of
in two ways:
Plugging in ,
, and
for
,
, and
respectively, and solving for
, we obtain
.
Then continue as before to use the Pythagorean Theorem on , find
, and find the volume of the pyramid.
Another Shortcut (Extended Law of Sines)
Take the base , where
and
. Draw an altitude from
to
that bisects
at point
. Then the altitude has length
. Next, let
. Then from the right triangle
,
. From the extended law of sines, the circumradius is
.
Solution 2 (Coordinates)
We can place a three dimensional coordinate system on this pyramid. WLOG assume the vertex across from the line that has length is at the origin, or
. Then, the two other vertices can be
and
. Let the fourth vertex have coordinates of
. We have the following
equations from the distance formula.
Adding the last two equations and substituting in the first equation, we get that . If you drew a good diagram, it should be obvious that
. Now, solving for
, we get that
. So, the height of the pyramid is
. The base is equal to the area of the triangle, which is
. The volume is
. Thus, the answer is
.
-RootThreeOverTwo
Solution 3 (Heron's Formula)
Label the four vertices of the tetrahedron and the midpoint of , and notice that the area of the base of the tetrahedron,
, equals
, according to Solution 1.
Notice that the altitude of from
to point
is the height of the tetrahedron. Side
is can be found using the Pythagorean Theorem on
, giving us
Using Heron's Formula, the area of can be written as
Notice that both and
can be rewritten as differences of squares; thus, the expression can be written as
From this, we can determine the height of both and tetrahedron
to be
; therefore, the volume of the tetrahedron equals
; thus,
-dzhou100
Solution 4 (Symmetry)
Notation is shown on diagram.
vladimir.shelomovskii@gmail.com, vvsss
Video Solution
https://youtu.be/Mk-MCeVjSGc ~Shreyas S
See Also
2017 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.